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If the electric flux entering and leaving an enclosed surface respectively are $\phi$, and $\phi_2$, the electric charge inside the surface will be
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Verified Answer
The correct answer is:
$\varepsilon_0\left(\phi_1+\phi_2\right)$
According to Gauss theorem "the net electric flux through any closed surface is equal to the net charge inside the surface divided by $\varepsilon_0{ }^{\prime \prime}$.
Therefore, $\quad \phi \propto \frac{q}{\varepsilon_0}$
Let $-q_1$ be the charge, due to which flux $\phi_1$ is entering the surface,
$\phi_1=\frac{-q_1}{\varepsilon_0}$
$-q_1=\varepsilon_0 \phi_1$
Let $+q_2$ be the charge, due to which flux $\phi_2$ is entering the surface.
$\phi_2=\frac{q_2}{\varepsilon_0}$
or $\quad q_2=\varepsilon_0 \phi_2$
So, electric charge inside the surface
$=q_2-q_1$
$\begin{aligned} & =\varepsilon_0 \phi_2+\varepsilon_0 \phi_1 \\ & \Rightarrow \quad=\varepsilon_0\left(\phi_2+\phi_1\right) \end{aligned}$
Therefore, $\quad \phi \propto \frac{q}{\varepsilon_0}$
Let $-q_1$ be the charge, due to which flux $\phi_1$ is entering the surface,
$\phi_1=\frac{-q_1}{\varepsilon_0}$
$-q_1=\varepsilon_0 \phi_1$
Let $+q_2$ be the charge, due to which flux $\phi_2$ is entering the surface.
$\phi_2=\frac{q_2}{\varepsilon_0}$
or $\quad q_2=\varepsilon_0 \phi_2$
So, electric charge inside the surface
$=q_2-q_1$
$\begin{aligned} & =\varepsilon_0 \phi_2+\varepsilon_0 \phi_1 \\ & \Rightarrow \quad=\varepsilon_0\left(\phi_2+\phi_1\right) \end{aligned}$
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